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Question 2 MCQ · 3.0 marks
Which one of the following options is true for the relation
?
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is a reflexive relation.
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is a symmetric relation.
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is a transitive relation.
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is an equivalence relation.
Published solution
**1. Test symmetry:** The condition \(x^2+y^2=1\) is unchanged when x and y are interchanged. Thus xRy implies yRx.
**2. Disprove reflexivity:** Reflexivity on the real numbers would require \(2x^2=1\) for every real x. It fails at x=0.
**3. Disprove transitivity:** 1R0 and 0R1 hold, but 1R1 fails because \(1^2+1^2=2\ne1\). Therefore the relation is symmetric but is neither reflexive nor transitive, and is not an equivalence relation.
**Answer: B.**
A: **Incorrect:** The relation fails reflexivity at zero.
B: **Correct:** The defining equation is invariant under swapping the two inputs.
C: **Incorrect:** The chain 1R0 and 0R1 does not imply 1R1.
D: **Incorrect:** Reflexivity and transitivity are required for equivalence and both fail.
**1. Read the slope:** \(2x+y+3=0\) is \(y=-2x-3\), with slope -2.
**2. Keep that slope for a parallel line:** Through (-1,1), the required line is \(y-1=-2(x+1)\), or \(y=-2x-1\).
**3. Substitute the other point:** At x=1, \(k=-2(1)-1=-3\).
**Answer: -3.**
be the largest possible interval such that the quadratic expression
is
positive for all
. Find
.
Published solution
**1. Factor:** \(-x^2+x+6=-(x-3)(x+2)\), with roots -2 and 3.
**2. Find the positive interval:** The leading coefficient is negative, so the quadratic is positive strictly between its roots. The largest open interval is (-2,3).
**3. Compute its length:** \(b-a=3-(-2)=5\).
**Answer: 5.**
Consider the following weighted undirected graph with weights written down next to the edges.
In order to obtain a minimum-cost spanning tree using Kruskal's algorithm, we start with the
graph with vertex set {A,B,C,D,E} and empty edge set. In the first step, we add (D,E), the edge with
the least weight. In the second step, we add the edge (D,C). What is the weight of the edge to be
added in the third step?
Published solution
**1. Follow Kruskal’s rule:** Examine edges by increasing weight, adding an edge only when it does not form a cycle.
**2. Track the first additions:** DE=6 and DC=10 connect vertices D,E,C. The next candidate CE=12 would complete the triangle C-D-E-C, so it is skipped.
**3. Add the next valid edge:** AB=15 connects two previously separate vertices and creates no cycle. Its weight is 15. Kruskal can grow several components before joining them; the new edge need not touch the existing C-D-E component.
**Answer: 15.**
**1. Focus on the leading term:** For large absolute x, \(p(x)\) has the sign and growth of \(a_nx^n\).
**2. Examine the left end:** As \(x\to-\infty\), odd powers are negative and even powers are positive.
**3. Obtain positive infinity:** A negative leading coefficient with odd n makes the result positive; a positive coefficient with even n does so too. Thus the pairs (-1,5) and (1,4) are selected.
**Answer: B, C.**
A: **Incorrect:** Positive coefficient with odd degree tends to negative infinity.
B: **Correct:** Negative coefficient times an odd power tending to negative infinity gives positive infinity.
C: **Correct:** Positive coefficient and even degree give positive infinity.
D: **Incorrect:** Negative coefficient and even degree give negative infinity.
be functions. Which of the following conditions are sufficient to
ensure that
is well-defined?
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Range
Domain
.
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Range
Domain
.
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Domain
Range
.
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.
Published solution
**1. Trace the composition:** \((g\circ f)(x)=g(f(x))\). The value f(x) must therefore be an admissible input to g.
**2. State the condition:** \[\operatorname{Range}(f)\subseteq\operatorname{Domain}(g).\] This guarantees g can be applied for every x in f’s domain.
**3. Distinguish the reverse condition:** Range(g) inside Domain(f) concerns the opposite composition. Also f’s actual range can be smaller than its declared codomain B, so requiring only C inside B does not guarantee f’s values belong to C.
**Answer: B.**
A: **Incorrect:** This addresses applying f after g, not g after f.
B: **Correct:** Every output of f is then a valid input to g.
C: **Incorrect:** The domain of f need not contain or be contained in g’s range for this composition.
D: **Incorrect:** C being contained in B is the wrong containment direction and need not cover f’s actual outputs.
for which partial information about the adjacency
matrix
is known as given below. An asterisk (*) denotes that the value is unknown.
.
Let the
entry of
be denoted by
. Which of the following options is true?
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There is a path of length two from the vertex
to the vertex
irrespective of
the value of
.
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There is a path of length two from the vertex
to the vertex
only if
.
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There is a path of length two from the vertex
to the vertex
only if
.
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The given data is insufficient to deduce the conditions under which a path of
length two from vertex
to vertex
exists.
Published solution
**1. Read known edges:** The displayed matrix has \(A[0,1]=1\) and \(A[1,2]=1\).
**2. Construct a two-edge path:** These entries give \(0\to1\to2\), regardless of any unknown edge from 0 to 3.
**3. Explain sufficiency:** One known intermediate vertex is enough to prove existence. Unknown entries can add other paths but cannot remove the displayed path.
**Answer: A.**
A: **Correct:** The fixed edges through vertex 1 establish the required path independently of A[0,3].
B: **Incorrect:** No condition A[0,3]=1 is needed.
C: **Incorrect:** No condition A[0,3]=0 is needed either.
D: **Incorrect:** The two known adjacency entries already settle existence.
**1. Check the bases:** The expression requires x>6 and x-6 not equal to 1, so x cannot be 7.
**2. Simplify logarithms:** Let \(u=\log_x3\), \(v=\log_{x-6}3\). Then \(\tfrac12\log_x9=u\) and \(\log_{x-6}27=3v\). The equation becomes \(u+v=3uv\).
**3. Divide and solve:** \(1/u+1/v=3\) gives \(\log_3x+\log_3(x-6)=3\), hence \(x(x-6)=27\). Solving \(x^2-6x-27=0\) gives 9 and -3. Only 9 satisfies the domain.
**Answer: 9.**
**1. Identify the leading growth:** \(a_n=[2n+5(-1)^n]/[4n-3(-1)^n]\). The oscillating terms remain bounded while n grows.
**2. Divide by n:** \[a_n=\frac{2+5(-1)^n/n}{4-3(-1)^n/n}.\]
**3. Take limits:** Both oscillating fractions tend to zero, so \(a_n\to2/4=0.5\). Oscillation does not prevent convergence when its size is divided by n.
**Answer: 0.5.**
**1. Check the indeterminate forms:** L’Hospital’s rule applies directly to ratios of type 0/0 or infinity/infinity under its differentiability and derivative-limit conditions.
**2. Examine each ratio:** \(\sin x/x\) is 0/0; differentiating gives \(\cos x/1\to1\). The ratio \((1-\cos x)/(1-e^x)\) is also 0/0; differentiating gives \(\sin x/(-e^x)\to0\).
**3. Exclude the others:** \(\cos x/x\) has numerator tending to 1. For \(\sin(1/x)/(1/x)\), the numerator does not have an infinite limit and is only bounded/oscillatory, so direct L’Hospital is not justified. That limit is instead zero by \(|x\sin(1/x)|\le|x|\).
**Answer: B, D.**
A: **Incorrect:** The ratio tends from 1/0, not a qualifying indeterminate form.
B: **Correct:** The 0/0 form satisfies the rule and gives limit 1.
C: **Incorrect:** Bounded oscillation over an unbounded denominator is not a direct infinity/infinity form.
D: **Correct:** The 0/0 form gives derivative ratio tending to zero.
Every topological sort of the graph must start with the vertex A.
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Every topological sort of the graph must end with the vertex E.
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ACBDE is a valid topological sort of the graph.
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ADCBE is a valid topological sort of the graph.
Published solution
**1. List precedence constraints:** The arrows are A→C, A→D, B→D, C→E and D→E. A topological ordering must place every source of an edge before its destination.
**2. Identify first/last possibilities:** Both A and B initially have indegree zero, so an order need not start with A. Every other vertex precedes E directly or through a path, so E must be last.
**3. Test the displayed orders:** ACBDE respects every arrow. ADCBE places D before B even though B→D, so it is invalid.
**Answer: B, C — Selected: Every topological sort of the graph must end with the vertex E.; ACBDE is a valid topological sort of the graph.**
A: **Incorrect:** B can be the first vertex, for example in BACDE.
B: **Correct:** All other vertices must precede E.
C: **Correct:** ACBDE satisfies each of the five precedence constraints.
D: **Incorrect:** ADCBE violates the B-before-D requirement.
The power consumption (in kW) of a household as a function of time (in hours past midnight) on a
particular day is given by the function
Answer the subquestions using the above function.
Find the maximum value of power consumption by the household in the time interval
.
Published solution
**1. Compare the two pieces:** On 0≤t<6, power is 0.25 kW. On 6≤t≤10, power is \(1.25+\sin[\frac\pi2(t-7)]\).
**2. Find the sine maximum:** Sine is at most 1. It reaches 1 when \(\frac\pi2(t-7)=\pi/2\), giving t=8, which is in the interval.
**3. Calculate:** The maximum is \(1.25+1=2.25\) kW, exceeding the earlier constant power.
**Answer: 2.25.**
The power consumption (in kW) of a household as a function of time (in hours past midnight) on a
particular day is given by the function
Answer the subquestions using the above function.
Find the number of units of electricity (1 unit = 1 kWh) consumed by the household in the time
interval
.
(Note that the electrical energy is the integral of the power function.)
Published solution
**1. Integrate each segment:** \[E=\int_0^6 0.25\,dt+\int_6^{10}\left[1.25+\sin\left(\frac\pi2(t-7)\right)\right]dt.\]
**2. Evaluate constant contributions:** These are \(0.25(6)=1.5\) kWh and \(1.25(4)=5\) kWh.
**3. Evaluate the sine contribution:** Its antiderivative is \(-\frac2\pi\cos[\frac\pi2(t-7)]\). At the two limits the arguments are -pi/2 and 3pi/2; both cosines are zero, so the net contribution is zero.
**4. Add:** Energy is \(1.5+5=6.5\) kWh, or 6.5 electricity units.
**Answer: 6.5.**
, then find the
cardinality of the maximum independent set of the graph
.
Published solution
**1. Define an independent set:** It contains vertices with no edges between any two of them. Number rows/columns of the given matrix 1 through 7.
**2. Exhibit five independent vertices:** The set \(\{1,3,5,6,7\}\) has zero adjacency entries between every pair. It therefore proves that the maximum size is at least five.
**3. Prove five is maximal:** The graph contains the disjoint edges (1,2) and (4,5). Any independent set must omit at least one endpoint from each edge, hence omit at least two distinct vertices out of seven. Thus its size is at most \(7-2=5\), matching the constructed set.
**Answer: 5.**
